Question-189140

Question Number 189140 by 073 last updated on 12/Mar/23

Answered by Ar Brandon last updated on 12/Mar/23

x>0 ∧ x≥1 ⇒x≥1 4−lnx≥3(√(lnx)) , t^2 =lnx 4−t^2 ≥3t ⇒ t^2 +3t−4 ≤ 0 (t+4)(t−1)≤0 ⇒−4≤t≤1 0≤t^2 ≤1 ⇒0≤lnx≤1 1≤x≤e ⇒x ∈ [1, e] ⇒b−a=e−1

$${x}>\mathrm{0}\:\wedge\:{x}\geqslant\mathrm{1}\:\Rightarrow{x}\geqslant\mathrm{1} \\ $$$$\mathrm{4}−\mathrm{ln}{x}\geqslant\mathrm{3}\sqrt{\mathrm{ln}{x}}\:,\:{t}^{\mathrm{2}} =\mathrm{ln}{x} \\ $$$$\mathrm{4}−{t}^{\mathrm{2}} \geqslant\mathrm{3}{t}\:\Rightarrow\:{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{4}\:\leqslant\:\mathrm{0} \\ $$$$\left({t}+\mathrm{4}\right)\left({t}−\mathrm{1}\right)\leqslant\mathrm{0}\:\Rightarrow−\mathrm{4}\leqslant{t}\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant{t}^{\mathrm{2}} \leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant\mathrm{ln}{x}\leqslant\mathrm{1} \\ $$$$\mathrm{1}\leqslant{x}\leqslant{e}\:\Rightarrow{x}\:\in\:\left[\mathrm{1},\:{e}\right]\: \\ $$$$\Rightarrow{b}−{a}={e}−\mathrm{1} \\ $$

Commented by manxsol last updated on 12/Mar/23

Sr.Ar Brandon ,con todo respeto. exercise=log(x) base 10 you exercise ln(x) base e

$${Sr}.{Ar}\:{Brandon}\:,{con}\:{todo}\:{respeto}. \\ $$$${exercise}={log}\left({x}\right)\:\:\:\:{base}\:\mathrm{10} \\ $$$${you}\:{exercise}\:{ln}\left({x}\right)\:\:\:\:{base}\:{e} \\ $$

Commented by Ar Brandon last updated on 12/Mar/23

Hola sen^ or ! log(x) is not necessarily to base 10. It all depends on your education system. In some countries logx implies to the base 10 while in others it is to the base e. By the way in most programming languages like the C programming language the function log(x) is to the base e. And to use the base 10 instead we must precise log10(x). Whatever be the case the logic applied in the above solution remains same. Gracias por prestar atencio^� n a mi solucio^� n.

$$\mathrm{Hola}\:\mathrm{se}\overset{ } {\mathrm{n}or}\:! \\ $$$$\mathrm{log}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{necessarily}\:\mathrm{to}\:\mathrm{base}\:\mathrm{10}.\:\mathrm{It}\:\mathrm{all}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{your}\:\mathrm{education} \\ $$$$\mathrm{system}.\:\mathrm{In}\:\mathrm{some}\:\mathrm{countries}\:\mathrm{log}{x}\:\mathrm{implies}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}\:\mathrm{10}\:\mathrm{while}\:\mathrm{in}\:\mathrm{others}\:\mathrm{it} \\ $$$$\mathrm{is}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}\:{e}.\:\mathrm{By}\:\mathrm{the}\:\mathrm{way}\:\mathrm{in}\:\mathrm{most}\:\mathrm{programming}\:\mathrm{languages}\:\mathrm{like}\:\mathrm{the}\:\mathrm{C} \\ $$$$\mathrm{programming}\:\mathrm{language}\:\mathrm{the}\:\mathrm{function}\:\mathrm{log}\left({x}\right)\:\mathrm{is}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}\:{e}.\:\mathrm{And}\:\mathrm{to}\:\mathrm{use}\: \\ $$$$\mathrm{the}\:\mathrm{base}\:\mathrm{10}\:\mathrm{instead}\:\mathrm{we}\:\mathrm{must}\:\mathrm{precise}\:\mathrm{log10}\left({x}\right). \\ $$$$\mathrm{Whatever}\:\mathrm{be}\:\mathrm{the}\:\mathrm{case}\:\mathrm{the}\:\mathrm{logic}\:\mathrm{applied}\:\mathrm{in}\:\mathrm{the}\:\mathrm{above}\:\mathrm{solution}\:\mathrm{remains}\:\mathrm{same}. \\ $$$$\mathrm{Gracias}\:\mathrm{por}\:\mathrm{prestar}\:\mathrm{atenci}\acute {\mathrm{o}n}\:\mathrm{a}\:\mathrm{mi}\:\mathrm{soluci}\acute {\mathrm{o}n}. \\ $$

Answered by mehdee42 last updated on 13/Mar/23

A)x>0 B)logx≥0⇒x≥10 C)((√(logx))+4)((√(logx))−1)≤0⇒ (√(logx))−1≤0⇒x≤10 ∩A,B,C=[1,10]⇒b−a=9

$$\left.{A}\right){x}>\mathrm{0} \\ $$$$\left.{B}\right){logx}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{10} \\ $$$$\left.{C}\right)\left(\sqrt{{logx}}+\mathrm{4}\right)\left(\sqrt{{logx}}−\mathrm{1}\right)\leqslant\mathrm{0}\Rightarrow \\ $$$$\sqrt{{logx}}−\mathrm{1}\leqslant\mathrm{0}\Rightarrow{x}\leqslant\mathrm{10} \\ $$$$\cap{A},{B},{C}=\left[\mathrm{1},\mathrm{10}\right]\Rightarrow{b}−{a}=\mathrm{9} \\ $$

Commented by manolex last updated on 13/Mar/23

$$\left.{B}\right)\:\:{x}\gg\mathrm{1} \\ $$

Commented by 073 last updated on 13/Mar/23

$$\mathrm{thanks}\:\mathrm{alot} \\ $$$$ \\ $$

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